Mahler · no prize · open since 1968

Mahler's 3/2 problem

Take a positive number and multiply it by 3/2, again and again. Can its fractional part, the part after the decimal point, stay below one half every time, for ever? Kurt Mahler asked in 1968 whether any number does this. He called such a number a Z-number and guessed that none exists. Nobody has proved him right or wrong. There is no prize, which may be why so few people have dug here. It is the arithmetic sibling of Rule 30's question: a free side and a constrained side that must agree for ever.

Seen

Integer part n

Every unit interval [n, n+1) holds at most one possible Z-number. For your n the picture follows a candidate ξ* chosen exactly from the last window of survivors (the middle of it). The dots are the fractional parts of ξ*·(3/2)^j, step by step. They stay below the dashed line at ½ as long as any candidate in [n, n+1) survives, and then one jumps above. The top strip is the parity of the integer parts. For n = 4 survivors last through j = 11; the longest below 2⁶³ last through j = 156.

Stated

The exact form (proved)

Write ξ·(3/2)^j = n_j + u_j, with n_j an integer and u_j in [0, ½). Then the integer parts follow one fixed map, n_j+1 = ⌈3n_j/2⌉. The fractional parts are fixed by the future parities b_k = n_k mod 2:

u_j = Σ_k≥0 b_j+k · 2^k / 3^k+1

So a Z-number is a positive integer whose orbit under ⌈3x/2⌉ keeps every one of those sums below ½. Read on its own, the fractional side is completely free: every point of [0, ½) has an infinite admissible future. The whole difficulty is the integer side, whether some positive whole number can produce an admissible record. The same series read in the 2-adic numbers gives −n₀. The 2-adic starts with an admissible record form a set of dimension log₂(3/2). Mahler's conjecture says that set contains no positive ordinary integer.

Known in print

No short window. Flatto, Lagarias and Pollington (1995): for every ξ > 0, the lim sup minus the lim inf of the fractional parts of ξ(3/2)ⁿ is at least 1/3, so no ξ keeps them in any interval shorter than 1/3. Windows. Dubickas (2008) excludes a union of two intervals and shows that infinitely many ξ stay inside (5/48, 43/48). Stephan (2026), in Lean, shows no ξ stays in (2/7, 5/7), a window of length 3/7. Computation. Dubickas and Mossinghoff (2009): a Z-number, if one exists, exceeds 2⁵⁷. Conditional. Andrieu, Eliahou and Vivion (2025): if every orbit is evenly spread modulo every power of two, there is no Z-number.

What this project added (October 2026)

A census to 2⁶³, both ways (computed, exact): no Z-number below 2⁶³, and the negative half, a number whose fractional parts all stay in (½, 1), is clear to the same size. One laptop, a quarter of an hour each. The census in full.

A ladder of residue problems (proved, with a second reader). For each g ≥ 3 and the windows [0, a) with a between 3^g−1/(3^g − 2^g) and (1 + (2/3)^g−1)/3, the fractional side drops out entirely. A number keeps all its fractional parts in such a window exactly when it is n + u₀(n) for a positive integer n whose odd orbit terms are all ≡ −1/3 modulo 2^g. The first rung, g = 3, reads: is there a positive integer whose odd orbit terms are all 5 mod 8? That is equivalent to a ξ with every fractional part below 9/19. These windows are wider than 1/3 and beyond the published exclusions, and their emptiness is open. Whether the equivalence itself is already in print we have not settled.

Two cautions for anyone attempting a proof (single-party, not yet second-read). An argument that never uses the sign of the start would also settle the open window (½, 1). And several standard integer-side tools see only that the start is rational, not that it is a whole number; on their own they would have to prove a strictly stronger conjecture about rational starts.

From outside: machine-checked in Lean (19 October 2026)

Joshua Coon, with the Astra and Argon AI models in GitHub Copilot, sent us a library of 142 Lean 4 files on this problem: every theorem checked by a machine, from the exact reduction to new windows wider than 1/3 that no number can stay in. We rebuilt it ourselves and recomputed its numbers; it checks out. It also checks, from outside, two of our own results: that Mahler's conjecture holds exactly when every bounded range of starts dies out, and the inequalities behind the ladder above. Its windows, drawn and checked live · the full entry.

Sources. MAHLER.md (the exact form, the census, the swarm's results and the ladder's proof); RULE30-GPT.md G50 (the exact form), GC1221 and G293 (the dimension), GC1228 and G294 (the ladder, second-read); PRIOR-ART.md ("Mahler's 3/2 problem: survey at the room switch"); K. Mahler, J. Austral. Math. Soc. 8 (1968); Flatto, Lagarias and Pollington, Acta Arith. 70 (1995); Dubickas, Math. Nachr. 281 (2008); Dubickas and Mossinghoff, Math. Comp. 78 (2009); Stephan (2026), Lean confinement certificates.

Checked, here, now: the first rung

Is there a positive integer whose odd orbit terms under ⌈3x/2⌉ are all 5 mod 8? If there is, a number exists with every fractional part of ξ(3/2)ⁿ below 9/19, and so a Z-number too. Type a start and see how long it plays the game.

Every start fails sooner or later in every search so far. Among the starts below 2⁶³ still alive after 124 steps, the share whose first 124 parity bits keep every two odd steps at least 3 apart (the first rung's pattern, which certifies the 5 mod 8 game through step 121) was what chance predicts: 217 of 4,499, against 214.

From outside: windows wider than 1/3

Joshua Coon's Lean library, built with the Astra and Argon AI models in GitHub Copilot, proves that no positive number keeps every fractional part of ξ·(3/2)ⁿ inside any of the sets below. Each is wider than Flatto, Lagarias and Pollington's 1/3, and each comes with a deadline. Pick a set and a size W. This page then follows every number between 1 and 2^W exactly, in your browser, and finds the step by which all of them have left.

W 6

For Mahler's own half-window [0, ½) the same check also ends, at step 4, 12, 25 and 29 as W grows from 1 to 8, and our census takes it to 2⁶³. The difference is the deadline: for each set above, Lean proves one for every W, while for [0, ½) nobody can.

A fractal in the middle window

Stephan (2026) proved in Lean that no number stays in the middle window (2/7, 5/7). Watch why it is so tight: follow every possible carry from step to step, and count how many separate branches of survivors remain. The library proves that the count equals Mahler's own binary partition numbers 1, 2, 4, 6, 10, 14, … (each is the one before plus the one halfway back) for 15 steps, and then falls behind. This page recomputes the counts exactly.

Dots: surviving branches in (2/7, 5/7) at each depth, computed exactly in your browser. Line: Mahler's binary partition numbers B(d) = B(d − 1) + B(⌊d/2⌋). They agree through depth 15; at depth 16 the window keeps 200 branches against 202. Hover a dot for its values.

Depth	Branches in (2/7, 5/7)	Mahler's B(d)	Difference

Sources. Joshua Coon, with Astra and Argon via GitHub Copilot, mahler-3-2-lean4 (the Dubickas extension module; checked here at f6d9492, see Outside contributions); Dubickas, Math. Nachr. 281 (2008), the union (8/39, 18/39) ∪ (21/39, 31/39); Stephan (2026), Lean confinement certificates; our recomputation, mahler_contrib_checks.py.

What it is and what it is not

No proof, and nothing claimed. The census is evidence and the ladder is a reformulation. Our own swarm of nine agents, run on this problem with a skeptic for every idea, found nothing that moves the critical path, and said so. What the problem needs is a reason why a whole number's finite binary expansion cannot produce an admissible record, and no such reason is known.

Where to go next

No Z-number below 2⁶³: the census, with a live exact checker and the cost of going further.
How the three problems connect, and its siblings Rule 30 and Collatz.
The open statements in symbols.
Outside contributions: Joshua Coon's Lean library, and how we checked it.